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when the output impedance seen by the filter is greater than the input impedance, it is best to use a T-configuration (series first) filter. No, this is not the case. In a 'normal' filter, both terminations are resistive, you can use either a Tee or a Pi configuration. When one of the terminations is designed to be either open circuit or short circuit, ...


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The MW supply will have a line filter on board. That said, there’s no harm in keeping the existing line filter. It won’t cause a malfunction and may actually help.


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OP oscillograph shows clipping of the bottom peak of the sinusoidal wave. This suggests that the internal buffer can only source current, and cannot sink current. A load resistor is required to pull the buffer output down. An internal pull-down resistor is included. From the data sheet spec, the 5k minimum pull-down resistor combined with \$V_{REFHI}\$ of +...


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Adding a high impedance low frequency Pole will degrade load regulation and add a high Q resonance at the pole frequency under low load conditions as the filter becomes underdamped. Load regulation from a Step load is defined by the ratio of source to load impedance divided by the loop gain spectrum. This shows up as overshoot and ripple spectrum. So very ...


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First, the filters have two different purposes, unless this were a true two-wire device, and the AOZ1284 is not. An input filter provides a smoothed voltage source to the buck converter. and it prevents high-frequency harmonics (RF noise) from getting into the supply line. An output filter eliminates ripple and noise from the output line. Second, the ...


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the antenna driver has a maximum at ~250mA which is well below the 4A rating of the inductor. Can a smaller inductor be chosen? A smaller inductor will almost certainly have a higher ESR. This will cause a higher DC voltage drop through the filter. If you can accept that limitation, you can probably use a smaller inductor. You should also consider whether ...


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If one was to design at the output a low-pass PI filter with a cutoff frequency of 10kHz for example (purely hypothetic), how would one proceed? You would proceed by going backwards and recognizing that you already have capacitance on the output of your current circuit so, all you actually might require is an LC low pass filter to tack on to the output. ...


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I would get Elsie, which even in the free Student Edition can create up to 7th order RLC filters. Full graphical design and the filters work. Using a combination of Elsie and balanced layout, I've found that designing and building filters from about 100Khz to over 470Mhz fairly easy. It goes without saying that the higher in frequency you go, the more ...


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Charge injection may or may not be a problem, depending on the performance required of the PLL. Using two devices switching in antiphase, and making the assumption that an on-going device generates the opposite charge injection to an off-going device, gives you a first order cancellation of the injection current. Uncancelled charge injection can be ...


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Q1 and the 1K resistor are mainly there in case \$Vdd > V_{in_{HI}}\$. The PMOS, Q2, requires: \$Vgs \ge 0\$ to turn off \$Vgs < 0 \$ to turn on In other words, drive the gate LO to turn it on and drive it HI to turn it off. But there are two problems if your try to drive Q2's gate directly with Vin while \$Vdd > V_{in_{HI}}\$: Vin cannot ...


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The input impedance of the micro is very high .No wonder there is a pronounced resonant peak as Mattman has shown .Damp the LC filter to get low Q and the nasty peak will go .Your coil should be screened like in a can so it wont pick up noise .Otherwise your proposed filter is valid .


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I used CircuitLab to plot the frequency response of the filter. I don't like it very much. Any noise at 10.5 kHz will hardly be filtered at all. It will filter fairly well above 15kHz. So, if that is where your noise is, this filter is OK. If your noise is lower than that, you should use a Sallen-Key butterworth filter. Or, if you show your entire circuit, ...


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I would recommended a circuit like this: If you plug in a DC adapter, the power will be drawn from the adapter. If no adapter is connected, the power will be drawn from the battery. To answer you other questions, you can't just charge any battery with another adapter. To filter noise, use suitable capacitors. Regarding the filtering, take a look for ...


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DC bias tolerance and temperature benefits smaller capacitors COG rated capacitance wont come in larger capacitance COG capacitor's capacitance change over time is negligible. DC bias is tighter for COG packages making them better suitable products for filtering applications higher value capacitors (uF and higher) can vary up-to 50% of the rated capacitor ...


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Smaller caps may be necessary for higher frequency or higher impedance or high voltage, but not always. Here, the application is not that high f, (as in RF) but the load impedance was stated to be high at 1 Meg. This makes it convenient to use small ceramic caps, which are inexpensive with a large R-value, where the breakpoint is defined by ω=1/RC. ...


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Basically, lower valued capacitors can use dielectric materials that have better performance. Less variability due to temperature and bias voltage. Less variability over time. Higher working voltage ratings. The best materials generally only deliver capacitance values up to 100 nF or so at reasonable prices. Here, we're talking about \$RC=1\ {\rm s}\$, so ...


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In summary, my original idea (using a crude constant-voltage power supply) will not work due to resistance in the circuit and motor. Using BEMF could actually reduce part count (one goal that I originally should have stated), and would actually work. Thank you DKNguyen, and Bruce Abbott for your thoughtful and thorough responses.


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Well, we have the following circuit: simulate this circuit – Schematic created using CircuitLab The input resistance is given by: $$\text{R}_\text{in}=\frac{\text{R}_1\left(\text{R}_2+\frac{\text{R}_3\text{R}_4}{\text{R}_3+\text{R}_4}\right)}{\text{R}_1+\text{R}_2+\frac{\text{R}_3\text{R}_4}{\text{R}_3+\text{R}_4}}\tag1$$ The input current is given ...


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Since I know that if you power a brushed motor with a constant-voltage power supply, it will always rotate at about the same speed, About the same speed, but not the same. In fact as you load the motor more its speed drops linearly, all the way to zero at stall. This is caused by the motor's internal resistances (brushes, armature windings etc.) that ...


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Your idea is wrong and won't work. But there is a very similar idea that will work. "a brushed motor with a constant-voltage power supply, it will always rotate at about the same speed" No. If you load it a lot, it will slow down a lot. Your idea is to basically monitor the average voltage across the motor terminals and adjust the PWM so that a constant ...


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for the first three questions uses Laplace transform, write the matrix equations, solve and the use the inverse Laplace transform


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